Q) Atal Tunnel : Atal Tunnel (also known as Rohtang Tunnel) is a highway tunnel built under the Rohtang Pass in the eastern Pir Panjal range of the Himalayas on the Leh-Manali Highway in Himachal Pradesh. At a length of 9.02 km, it is the longest tunnel above 10,000 feet (3,048 m) in the world and is named after former Prime Minister of India, Atal Bihari Vajpayee. The tunnel reduces the travel time and overall distance between Manali and Keylong on the way to Leh. Moreover, the tunnel bypasses most of the sites that were prone to road blockades, avalanches, and traffic snarls.
Earth is excavated to make a railway tunnel. The tunnel is a cylinder of radius 7 m and length 450 m. A level surface is laid inside the tunnel to carry the railway lines. Figure given below shows the circular cross – section of the tunnel. The level surface is represented by AB , the centre of the circle is O and ∠ AOB = 90c. The space below AB is filled with rubble (debris from the demolition buildings).
(i) How much volume of earth is removed to make the tunnel?
(ii) If the cost of excavation of 1 cubic meter is Rs 250, what is the total cost of excavation?
(iii) A coating is to be done on the surface of inner curved part of tunnel. What is the area of tunnel to be being coated ? Costing of coating is Rs 30 per m2 . What is the total cost of coating?
(iv) How much volume of debris is required to fill the ground surface of tunnel?
Ans:
(i) Earth removed to make tunnel:
Since the tunnel was made cylindrical by earth excavation, the earth removed was equal to the volume of the cylindrical tunnel.
Next, we know the volume of the Cylinder = π r2 h
here, r is the radius of the cylinder and h is the height of the cylinder
Next, we are given the radius of the tunnel, r = 7 m, and the length of the tunnel, h = 450 m
∴ the volume of the cylindrical tunnel = π r2 h = π (7)2 (450)
= 
x (7)2 (450) = 22 x 7 x 450
= 69,300 m3
Therefore, the earth removed to make the tunnel was 69,300 m3
(ii) Cost of excavation:
Now the cost of excavation = Volume of earth excavated x cost per m3
= 69,300 x 250 = 1,73,25,000
Therefore the cost of excavation is Rs. 1,73,25,000
(iii) Coating area and Coating cost:
(a) Coating area:
Since coating is to be done only on the curved surface area of the tunnel, we need to calculate total curved surface area and then remove the area under chord AB where rubble is filled.
Step 1: Now Curved surface area of a cylinder = 2 π r h,
here, r is the radius of the cylinder and h is the height of the cylinder
Next, we are given the radius of the tunnel, r = 7 m, and the length of the tunnel, h = 450 m
Total curved surface area of a cylinder = 2 x x (7) (450)
= 44 x 450 = 19,800 m2
Step 2: Let’s analyse the given circle diagram: 
Here, the minor segment (shown in grey color, under chord AB) is to be removed.
Length of the minor segment
= 2 π r (
)
and hence the area of tunnel due to the minor segment = 2 π r h ()
Given that the angle, θ is 900
the area of tunnel due to the minor segment = 2 x () (7) (450) (
)
= 2 x 22 x 450 x 
= 11 x 450 = 4,950 m2
Step 3: Next, we calculate the net curved surface area of the tunnel:
Net curved surface area of the tunnel for coating = Total curved surface area – area under chord AB (where rubble is filled)
= 19,800 – 4,950 = 14,850 m2
Therefore, the area of the tunnel to be coated is 14,850 m2
(b) Cost of coating:
Total Coating cost = Coating Area x Cost of coating per m2
= 14,850 x 30 = 4,45,500
Therefore the cost of coating is Rs. 4,45,000
(iv) Volume of Debris:
From the given diagram, we can see that the debris is to be filled in the area under chord AB (shown in grey color)
Area of a segment is given by:
π r 2 ()
And hence the volume of this segment along the length of the tunnel =
Area x length
= π r 2 () x h = π r 2 h ()
= () (7)2 (450) ()
= 22 x 7 x 450 x ()
= 11 x 7 x 225 = 11 x 1575 = 17325
Therefore, the volume of debris is 17,325 m3
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